[BZOJ 2738]矩阵乘法

来源：互联网发布：淘宝刷客被警察抓编辑：程序博客网时间：2024/05/11 04:06

卡时卡的我心塞啊

因为插入操作太多而且没有顺序我们需要在sort后的插入操作上滚来滚去= =

其实并不需要排序opt数组，而用int下标排序可避免复制的时间过长。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <ctime>#define maxn 510using namespace std;void read(int &num){num = 0;char ch = getchar();for(;ch < '!'; ch = getchar());for(;ch > '!'; ch = getchar())num = num * 10 + ch - 48;}int N, Q;//BIT{int t[maxn][maxn];int lowbit[maxn];#define lowbit_cal(x) (x & ((~x) + 1))void update(int x, int y, const int& val){if(!x || !y)return;int tmp = y;while(x <= N){y = tmp;while(y <= N){t[x][y] += val;y += lowbit_cal(y);}x += lowbit_cal(x);}}int ask(int x, int y){        if(!x || !y)return 0;int tmp = y, ans = 0;while(x){y = tmp;while(y){ans += t[x][y];y -= lowbit_cal(y);}x -= lowbit_cal(x);}return ans;}int ask(const int& x1, const int& y1, const int& x2, const int& y2){int ans = ask(x1 - 1, y1 - 1) + ask(x2, y2);ans -= ask(x2, y1 - 1) + ask(x1 - 1, y2);return ans;}int n;#define maxq 310010struct opt{int x1, y1, tp;int x2, y2, k;bool operator<(const opt& k)const{return tp < k.tp;}}q[maxq], q1[maxq], q2[maxq];const int inf = 0x7fffffff;int TMP, ans[maxq];int T = 0;void solve(int head, int tail, int l, int r){    //cout << head << ' ' << tail << ' ' << l << ' ' << r << endl;if(head > tail)return;if(l == r){for(int i = head; i <= tail; i ++)ans[q[i].tp] = l;return;}int mid = l + r >> 1;while(q[T + 1].tp <= mid && T < TMP)    T ++, update(q[T].x1, q[T].y1, 1);while(q[T].tp > mid && T)    update(q[T].x1, q[T].y1, -1), T --;int l1 = 0, l2 = 0;for(int i = head; i <= tail; i ++){if(ask(q[i].x1, q[i].y1, q[i].x2, q[i].y2) >= q[i].k)q1[++ l1] = q[i];else{//q[i].k -= tmp[i];q2[++ l2] = q[i];}}int cnt = head;for(int i = 1; i <= l1; i ++)q[cnt ++] = q1[i];for(int i = 1; i <= l2; i ++)q[cnt ++] = q2[i];//memcpy(q + head, q1 + 1, sizeof (q[0]) * l1);//memcpy(q + head + l1, q2 + 1, sizeof (q[0]) * l2);solve(head, head + l1 - 1, l, mid);solve(head + l1, tail, mid + 1, r);}int main(){    freopen("nt2012_mat.in","r",stdin);freopen("nt2012_mat.out","w",stdout);read(N);read(Q);TMP = N * N;int x;int mn = inf, mx = -inf;for(int i = 1; i <= N; i ++){for(int j = 1; j <= N; j ++){read(x);        if(x < mn)mn = x;if(x > mx)mx = x;++ n;q[n].x1 = i;q[n].y1 = j;q[n].tp = x;}}sort(q + 1, q + 1 + n);int id = 0;for(int i = 1; i <= Q; i ++){++ n;read(q[n].x1), read(q[n].y1);read(q[n].x2), read(q[n].y2);read(q[n].k);q[n].tp = ++ id;}solve(TMP + 1, n, mn, mx);   for(int i = 1; i <= Q; i ++)printf("%d\n", ans[i]);    return 0;}

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